I got it! I left out a times two...I was trying to be smooth with my algebra to make my calculations easier and left out a x 2. I got it. R=784624 so R+r=1.78562x10^6. Using the third equation that gives me 35 minutes. Thank you, at least I knew I was working it out correctly!
Homework Statement
A satellite is in circular orbit at an altitude of 1000 km above the surface of a nonrotating planet with n orbital speed of 5.3 km/s. The escape velocity for the planet is 11.3 km/s. In this situation the orbital period of the satellite, in minutes, is...?
Homework...
Homework Statement
What is the current and potential difference across the 4Ω, 12Ω, and 8Ω resistors.
What is the potential difference across a and b
https://fbcdn-sphotos-a.akamaihd.net/hphotos-ak-ash4/389632_3598331272733_1107039398_3285590_155442550_n.jpg
Homework Equations
V=IR...
I am really trying to understand what you are doing, but I haven't done three dimensional analysis before, I am only through Calculus 1...that was the only prereq. for this course, so bare with me...I assumed you named them respectively, so why is p parallel to 2i + 2j + k?
I sent my...
Can I do that? Can I replace ω with (v/r)? Is it the same velocity? So then my answer would be:
mgh=(1/2)Iω2 + (1/2)mv2
2450=v2(30)
v=9.04 m/s
ω=(9.04/.5)=18.07 rev/s?
This is pretty much a shot in the dark. I have never done anything in 3-dimensions We only covered this in class for a bit. The instructor tried to jam 3 chapters into one class, so here we go
a.
cos\theta=0
\theta=90
b.∑ τ =I\alpha
\tau=r x F sin \theta
I=(1/12)(50)(22+32)=54.17...
Does that mean I need to use the energy equation:
mgh=Iω2+.5mv2
To find v=at; have to find Xf=.5at2
t=2.26s; v=9.8(2.26)=22.15m/s
(10)(9.8)(25)=(12.5)ω2+.5(10)22.152
but that gives me -.16 for ω2??
Is my mgh correct?
Homework Equations
I=.5Mr2
Work=Fd
Force=ma
Energy: mgh=.5Iω2The Attempt at a Solution
Initial angular velocity is 0 rad/s
r=.5m
F=ma=(10)(9.8)=98N
W=(98)(25)=2450 J
I=(.5)(100)(.52)=12.5
2450=(.5)12.5ω2
ω=19.79rad/s---is that even close to right?
Homework Statement
a cylinder, r= 0.5 m with a mass 100 kg which is hanging 50 m above the ground. A rope of negligible mass which is 25 m long is wrapped around the cylinder. At the end of the rope a 10 kg mass is hanging. The hanging mass will fall and the rope will spin off the cylinder...
Homework Statement
A solid rectangle of uniform density has one corner at the origin. It has a mass of 50 kg. The rectangle
has a length of 4 m in the z-direction, a length of 3 m in the y-direction, and a length of 2 m in
the x-direction. The pivot is at the center of mass.
There is a 50 N...